If θi>0 for 1≤θ≤n and θ1+θ2+θ3+⋯+θn=π, then the greatest value of sum sin⁡θ1+sin⁡θ2+sin⁡θ3+⋯+sin⁡θn is equal to

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nsin⁡πn

None of these

answer is B.

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Detailed Solution

We have, 00,i=1,2,3,…n Now, sin⁡θ1+sin⁡θ2+sin⁡θ3+…+sin⁡θnn≤sin⁡θ1+θ2+θ3+…+θnn=sin⁡πn∴ sin⁡θ1+sin⁡θ2+sin⁡θ3+⋯+sin⁡θn≤nsin⁡πn

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